## Register Now

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi adipiscing gravdio, sit amet suscipit risus ultrices eu. Fusce viverra neque at purus laoreet consequa. Vivamus vulputate posuere nisl quis consequat.

## Lost Password

Lost your password? Please enter your username and email address. You will receive a link to create a new password via email.

## 7. The roots of each of the following quadratic equations are real and equal, find k. (1) 3$y^2$ + ky + l2 = 0 (2) kx (x – 2) + 6 = 0

Print or Save

(1)    3$y^2$ + ky + l2 = 0
Sol.  3$y^2$ + ky + l2 = 0
Comparing with a$x^2$ + bx + c = 0
we get, a = 3, b = k and c = 12

Let’s find value of discriminant
Δ   =   D  =   $b^2$ – 4ac
=   $k^2$ – 4 × 3 × 12
=   $k^2$ – 144                               ...(I)
but roots of equation are real and equal
∴     Δ  =   D    =   0                        ...(II)
∴     $k^2$ – 144  =   0               [From (I) & (II)]
∴     $k^2$ – 122  =   0
By using $a^2$$b^2$ = (a + b) (a – b)
∴   (k + 12)  (k – 12)  =  0
∴   k + 12  =  0    or   k – 12  =  0
∴          k   = –12 or          k   = 12

(2)   kx (x – 2) + 6 = 0
Sol. kx (x – 2) + 6 = 0
∴     k$x^2$ – 2kx + 6 = 0
comparing with a$x^2$ + bx + c = 0
∴    a = k, b = –2k and c = 6
Let’s find value of discriminant
Δ  =  D  =  $b^2$ – 4ac
=  $(–2k)^2$ – 4 × k × 6
=  4$k^2$ – 24k                             ...(I)
But roots of equation are real and equal
∴    Δ   =   D     =  0                                 ...(II)
∴    4$k^2$ – 24k  =  0                    [From (I) & (II)]
∴    4k(k – 6)    =  0
k – 6    =  $0\over 4k$
∴           k – 6    =  0
∴                 k    =  6

uÉaÉïxÉqÉÏMüUhÉ August 10 , 2018 0 Comments 135 views